Weak topology Banach space with separable dual

Question

Let $B$ be a Banach space with separable dual and let $(f_n)$ be dense and countable in $B^*$. Let $\tilde{\tau}$ be the initial topology associated to the collection of maps $f_n : B\rightarrow \mathbb{R}$.

My question: is $\tilde{\tau}$ the standard weak topology on $B$?

My attempt :

Let $\tau$ denote the weak topology on $B$. Obviously, $\tau$ makes all the $f_n$'s continuous. Being $\tilde{\tau}$ the smallest doing so, $$\tilde{\tau}\subseteq \tau.$$

Conversely, I tried to reason with the basis of such topologies. Fix arbitrary $x_0 \in B$, $\epsilon >0$ and $g_1,...,g_N \in B^*$ and recall that $U_{x_0}(\epsilon,g_1,...,g_N):= \{x \in B \colon |g_i(x-x_0)|< \epsilon, \ i=1,...,N\}$ is open neighborhood of $x_0$ in $\tau$. To conclude, it is sufficient to show that there exists an open neighborhood $\tilde{U}$ of $x_0$ in $\tilde{\tau}$ so that $\tilde{U}\subset U_{x_0}(\epsilon,g_1,...,g_N)$.

My guess is to pay some $\tilde{\epsilon}$ in requiring $f_{n_i} \approx g_i$ for all $i=1,..,N$ and define $\tilde{U}:= U_{x_0}(\tilde{\epsilon},f_{n_1},...,f_{n_N})$, but I am struggling in bounding the term $|f_{n_i}(x)-g_i(x)|$ uniformly on $x$.

Answer 1

Contrary to what has been stated in another answer, the answer to your question is no. By the result in the answer to this question every $\tilde{\tau}$-continuous linear functional on $B$ belongs to ${\rm span} \, \{f_{n} : n\in\mathbb{N} \}$. However, if we had $X^{*} = {\rm span} \, \{f_{n} : n\in\mathbb{N} \}$ then $X^{*}$ would have a countable Hamel basis. This is only possible if $X$ is finite-dimensional by this result, so the topology $\tilde{\tau}$ will never coincide with $\tau$ whenever $X$ is infinite-dimensional.

Answer 2

If $(f_n)$ is supposed to be dense in the norm of $B^{*}$ then this is quite easy. Let $f \in B^{*}$. There exist $n_1<n_2<...$ such that $\|f_{n_i}-f\| \to 0$. This implies that $f_{n_i} \to f$ uniformly on any ball in $B$. Since each $f_{n_i}$ is continuous w.r.t. $\overline {\tau}$ it follows that $f$ is also continuous w.r.t $\overline {\tau}$. Thus every $f \in B^{*}$ is continuous w.r.t $\overline {\tau}$. Hence $\tau \subset \overline {\tau}$.